document.write( "Question 301258: A bag contains balls numbered 1,2,3 first a ball is drawn from the bag and a fair coin is tosed the number of times as the number shown on the drawn ball.find the mean of number of heads?
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Algebra.Com's Answer #216013 by Edwin McCravy(20060)\"\" \"About 
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document.write( "Mean number of heads = 0*P(0 heads) + 1*P(1 head) + 2*P(2 heads) + 3*P(3 heads).\r\n" );
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document.write( "We don't need the probability of 0 heads since that will be multiplied\r\n" );
document.write( "by 0. So:\r\n" );
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document.write( "Mean number of heads = 1*P(1 head) + 2*P(2 heads) + 3*P(3 heads).\r\n" );
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document.write( "The probability of getting x heads out of n tosses of a fair coin is\r\n" );
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document.write( "\"nCx%2A%281%2F2%29%5Ex%2A%281%2F2%29%5E%28n-x%29=nCx%2A%281%2F2%29%5En\"\r\n" );
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document.write( "Use that formula to fill in this table of probabilities:\r\n" );
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document.write( "                   P(1 head)    P(2 heads)    P(3 heads)\r\n" );
document.write( "ball#1(1 toss)         1/2           0             0      \r\n" );
document.write( "ball#2(2 tosses)       1/2          1/4            0\r\n" );
document.write( "ball#3(3 tosses)       3/8          3/8           1/8 \r\n" );
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document.write( "Now we need to get P(1 head), P(2 heads), P(3(heads)\r\n" );
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document.write( "P(1 head) = P[(ball#1 AND 1 head) OR (ball#2 AND 1 head) OR (ball#3 AND 1 head)]\r\n" );
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document.write( "Using the rule of \"AND implies multiplication\" and \"OR implies addition\":\r\n" );
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document.write( "P(1 head) = \r\n" );
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document.write( "P(2 heads) = P[(ball#2 AND 2 heads) OR (ball#3 AND 2 heads)] = \"%281%2F3%29%2A%281%2F4%29%2B%281%2F3%29%2A%283%2F8%29=1%2F12%2B1%2F8=5%2F24\"\r\n" );
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document.write( "P(3 heads) = P[ball#3 AND 3 heads] = \"%281%2F3%29%2A%281%2F8%29=1%2F24\"\r\n" );
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document.write( "So,\r\n" );
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document.write( "Mean number of heads = 1*P(1 head) + 2*P(2 heads) + 3*P(3 heads) = .\r\n" );
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document.write( "So, surprisingly, the mean number of heads is 1 head!!!\r\n" );
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document.write( "Edwin
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